Question

$N{H}_{2}COON{H}_4\left(s\right)$ on heating to at $200K$ gives a mixture of $N{H}_3$ and $C{O}_2$ vapours having vapour density 13. The degree of dissociation of $N{H}_{2}COON{H}_4$ is :

• A. $33.3$%
• B. $44.6$%
• C. $55.4$%
• D. $66.7$%

Answer 1

Answer: (A)

$33.3$%

Molar mass of the mixture is twice the vapor density.
$M=2\times 13=26g/mol$

Let the mole fraction of ammonia in the gaseous mixture be $x$.
The mole fraction of $C{O}_2$ will be $1-x$.

Hence, the average molar mass of the gaseous mixture is $\frac{17x+44(1-x)}{1}=26$
Hence, $x=0.67$

Thus, the mole fractions of ammonia and carbon dioxide are 0.67 and 0.33 respectively.
The partial pressures of ammonia and carbon dioxide will be 0.67 atm and 0.33 atm respectively.

The equilibrium constant is $K_p=P_{N{H}_3}^2{P}_{C{O}_2}=\left(0.67{)}^2\right(0.33)=\frac{4}{27}$

Let y be the degree of dissociation.
If originally, 1 mole of ammonium formate is present, then at equilibrium, 2y moles of ammonia and y moles of carbon dioxide will be formed.

The expression for the equilibrium constant is $K_p=(2y{)}^{2}y=4y^3=\frac{4}{27}$

Hence $y=\frac{1}{3}$

Hence, the degree of dissociation is 33.3 %.